Relational Formulation of Quantum Measurement

Abstract

Quantum measurement and quantum operation theory is developed here by taking the relational properties among quantum systems, instead of the independent properties of a quantum system, as the most fundamental elements. By studying how the relational probability amplitude matrix is transformed and how mutual information is exchanged during measurement, we derive the formulation that is mathematically equivalent to the traditional quantum measurement theory. More importantly, the formulation results in significant conceptual consequences. We show that for a given quantum system, it is possible to describe its time evolution without explicitly calling out a reference system. However, description of a quantum measurement must be explicitly relative. Traditional quantum mechanics assumes a super observer who can instantaneously know the measurement results from any location. For a composite system consists space-like separated subsystems, the assumption of super observer must be abandoned and the relational formulation of quantum measurement becomes necessary. This is confirmed in the resolution of EPR paradox. Information exchange is relative to a local observer in quantum mechanics. Different local observers can achieve consistent descriptions of a quantum system if they are synchronized on the information regarding outcomes from any measurement performed on the system. It is suggested that the synchronization of measurement results from different observers is a necessary step when combining quantum mechanics with the Relativity Theory.

Appendices

Appendix A: Theorem 1

Theorem 1

H(R) = 0 if and only if the matrix elementR_ijcan be decomposed asR_ij = c_id_j,wherec_iandd_jare complex numbers.

Proof

According to the singular value decomposition, the relational matrix R can be decomposed to R = UDV, where D is rectangular diagonal and both U and V are N × N and M × M unitary matrix, respectively. This gives ρ = RR^‡ = U(DD^‡)U^‡. If H(R) = 0, matrix ρ is a rank one matrix, therefore DD^‡ is diag{1, 0, 0...}. This means D is a rectangular diagonal matrix with only one eigenvalue e^iϕ. Expanding the matrix product R = UDV gives

$$ R_{ij}=\sum\limits_{nm}U_{in}D_{nm}V_{mj}=U_{i1}e^{i\phi}V_{1j}. $$

(A1)

We just choose c_i = U_i1 and d_j = e^iϕV_1j to get R_ij = c_id_j. Conversely, if R_ij = c_id_j, R can be written as outer product of two vectors,

$$ R = \left( \begin{array}{llll} c_{1} & c_{2} & {\ldots} & c_{n} \end{array}\right)^{T} \times \left( \begin{array}{llll} d_{1} & d_{2} & {\ldots} & d_{m} \end{array}\right). $$

(A2)

Considering vector C₁ = {c₁,c₂,…,c_n} as an eigenvector in Hilbert space $\mathcal {H}_{S}$, one can use the Gram-Schmidt procedure [19] to find orthogonal basis set C₂,…,C_n. Similarly, considering vector D₁ = {d₁,d₂,…,d_m} as an eigenvector in Hilbert space $\mathcal {H}_{A}$, one can find orthogonal basis set D₂,…,D_m. Under the new orthogonal eigenbasis, R becomes a rectangular diagonal matrix D = diag{1, 0, 0...}. Therefore R = UDV where U and V are two unitary matrices associated with the eigenbasis transformations. Then ρ = RR^‡ = U(DD^‡)U^‡, and DD^‡ = diag{1, 0, 0...} is a square diagonal matrix. Since the eigenvalues of similar matrices are the same, the eigenvalues of ρ are (1, 0, ...), thus H(R) = 0. □

Appendix B: Decomposition of the Unitary Operator of a Bipartite System

If there is interaction between S and A, and the initial state of S + A is a product state, the global unitary operator can be decomposed into a set of measurement operators that satisfies (16). The proof shown here closely follows idea from Ref. [22]. Denote the initial state is product state, |Ψ₀〉 = |ψ₀〉_S|ϕ₀〉_A. First we change the eigenbasis for A through a local unitary operator $I_{S}\otimes \hat {U}_{A}$ such that ϕ₀ is the first eigenvector of the orthogonal eigenbasis, i.e., $(I_{S}\otimes \hat {U}_{A})|\psi _{0}\rangle _{S} \otimes |\phi _{0}\rangle _{A} = |\psi _{0}\rangle _{S}\otimes |a_{0}\rangle _{A}$, and {|a_m〉} forms an orthogonal eigenbasis of A. The global unitary operator is changed to $\hat {U}_{SA}^{\prime } = \hat {U}_{SA}(I_{S}\otimes \hat {U}_{A}^{\dag })$. Define a linear operator $\hat {M}_{m} = \langle a_{m}|\hat {U}_{SA}^{\prime }| a_{0}\rangle$. The set of operators $\{\hat {M}_{m}\}$ is what we are looking for, since we can verify it satisfies (16),

$$\begin{array}{@{}rcl@{}} \hat{U}_{SA}|\psi_{0}\rangle|\phi_{0}\rangle & =&\hat{U}_{SA}^{\prime}|\psi_{0}\rangle|a_{0}\rangle \\ & =& \sum\limits_{m} |a_{m}\rangle\langle a_{m}| \hat{U}_{SA}^{\prime}|\psi_{0}\rangle|a_{0}\rangle \\ & =& \sum\limits_{m}\hat{M}_{m}|\psi_{0}\rangle|a_{m}\rangle. \end{array} $$

(B1)

The completeness condition can also be verified,

$$\begin{array}{@{}rcl@{}} \sum\limits_{m}\hat{M}_{m}^{\dag}\hat{M}_{m} & =& \sum\limits_{m}\langle a_{0}|\hat{U}_{SA}^{\prime\dag}|a_{m}\rangle\langle a_{m}|\hat{U}_{SA}^{\prime}|a_{0}\rangle \\ & =& \langle a_{0}|\hat{U}_{SA}^{'\dag}\hat{U}_{SA}^{\prime}|a_{0}\rangle = I_{S}. \end{array} $$

(B2)

Appendix C: Theorem 2

Theorem 2

Applying operator$\hat {Q}\otimes \hat {O}$overthe composite systemS + Ais equivalent to change the relational matrix R toR^′ = QRO^T,where the superscript T represents a transposition.

Proof

Denote the initial state vector of the composite system as $|{\Psi }_{0}\rangle ={\sum }_{ij}R_{ij}|s_{i}\rangle |a_{j}\rangle$. Apply the composite operator $\hat {Q}(t)\otimes \hat {O}(t)$ to the initial state,

$$\begin{array}{@{}rcl@{}} |{\Psi}_{1}\rangle & =& (\hat{Q}\otimes \hat{O}) \sum\limits_{ij} R_{ij} |s_{i}\rangle \otimes |a_{j}\rangle \\ & =& \sum\limits_{ij}R_{ij}\hat{Q}|s_{i}\rangle\otimes \hat{O}|a_{j}\rangle \\ & =& \sum\limits_{ij}\sum\limits_{mn}R_{ij}Q_{mi}O_{nj}|s_{m}\rangle\otimes |a_{n}\rangle \\ & =& \sum\limits_{mn}(\sum\limits_{ij}Q_{mi}R_{ij}O^{T}_{jn})|s_{m}\rangle\otimes |a_{n}\rangle \\ & =&\sum\limits_{mn}(QRO^{T})_{mn}|s_{m}\rangle|a_{n}\rangle \end{array} $$

(C1)

where T represents the transposition of matrix. Compared the above equation to (7) for the definition of |Ψ₁〉, it is clear that the relational matrix is changed to R^′ = QRO^T. □

Appendix D: Probability in Selective Measurement

Given the composite system S + A is described by (15), the reduced density matrix of S can be defined

$$ \begin{array}{ll} {\displaystyle}\hat{\rho}_{S} &{\displaystyle} =Tr_{A}|{\Psi}\rangle\langle{\Psi}| = \sum\limits_{ii^{\prime}}(\sum\limits_{k}R_{ik}R^{*}_{i^{\prime}k})|s_{i}\rangle\langle s_{i^{\prime}}| \\ &{\displaystyle} =\sum\limits_{ii^{\prime}}(RR^{\dag})_{ii^{\prime}}|s_{i}\rangle\langle s_{i^{\prime}}| \end{array} $$

(D1)

and the probability of finding event |s_i〉 occurred to S is calculated by (10). Similarly, the probability of event |a_j〉 occurred to A is ${p^{A}_{j}}={\sum }_{i}p_{ij}={\sum }_{i}|R_{ij}|^{2}$. This can be more elegantly written by introducing a partial projection operator $I^{S}\otimes \hat {P}_{j}^{A}$ where $\hat {P}_{j}^{A}=|a_{j}\rangle \langle a_{j}|$. It is easy to verify that

$$\begin{array}{@{}rcl@{}} {p_{j}^{A}} & =&\langle{\Psi}|I^{S}\otimes \hat{P}_{j}^{A}|{\Psi}\rangle \\ & =&\langle{\Psi}|a_{j}\rangle\langle a_{j}|{\Psi}\rangle =\sum\limits_{i}|R_{ij}|^{2}. \end{array} $$

(D2)

Appendix E: Open Quantum System

The measurement theory described in this paper is consistent with the open quantum system (OQS) theory. OQS studies the dynamics when a quantum system interacts with its environment E [19, 24, 25]. Such interaction can result in entanglement and information exchange between the quantum system S and the environment system E. Recall the definition of apparatus in Section 2.1 includes the interacting environment as one type of apparatus, if we replace the environment system E with the apparatus system A, the OQS theory gives the same formulations as shown in Section 3.

First, we give a brief review of the OQS theory. Suppose the initial composite state for the quantum system and environment is described by a density matrix ρ_SE, the interaction between S and E changes the density matrix $\rho _{SE} \rightarrow \hat {U}\rho _{SE}\hat {U}^{\dag }$. The resulting density matrix of S is $\rho ^{\prime }_{S} = Tr_{E}(\hat {U}\rho _{SE}\hat {U}^{\dag })$. Denote the orthogonal eigen basis of the environment {|e_k〉}. Since the orthogonal eigenbasis of the environment is not necessary the same eigenbasis that diagonalizes ρ_E, we further denote the spectral decomposition of ρ_E as $\rho _{E} ={\sum }_{m}\lambda _{m}|\tilde {e}_{m}\rangle \langle \tilde {e}_{m}|$. Assumed the initial state of the quantum system plus environment is a product state, i.e., ρ_SE = ρ_S ⊗ ρ_E, the density operator of S after the interaction with the environment is

$$\begin{array}{@{}rcl@{}} \rho^{\prime}_{S} & =& {\Lambda} (\rho_{S}) \\ & =& \sum\limits_{mk}|\lambda_{m}|^{2} \langle e_{k}| \hat{U}(\rho_{S}\otimes |\tilde{e}_{m}\rangle\langle \tilde{e}_{m}| )\hat{U}^{\dag}|e_{k}\rangle \\ & =& \sum\limits_{mk}E_{mk}\rho_{S} E_{mk}^{\dag}. \end{array} $$

(E1)

where $E_{mk}=\lambda _{m}\langle e_{k}| \hat {U}|\tilde {e}_{m}\rangle$ and satisfies the completeness condition ${\sum }_{mk}E_{mk}E_{mk}^{\dag } =I$. Equation (E1) is the Kraus representation of the linear map Λ. It is proved that Λ can be a Kraus representation if and only if it can be induced from an extended system with initial condition ρ_SE = ρ_S ⊗ ρ_E [25]. If ρ_E = |e₀〉〈e₀| is a pure state, the linear map is further simplified to ${\Lambda } (\rho _{S})={\sum }_{k}E_{k}\rho _{S} E_{k}^{\dag }$ and $E_{k}=\langle e_{k}| \hat {U}|e_{0}\rangle$. The operator set {E_k} forms a POVM, and Λ(ρ_S) is a Complete Positive Trace Preserving (CPTP) map [21]. To connect to the measurement theory, suppose the measurement outcome m corresponds to an orthogonal state |ϕ_m〉 of E, and represented by a projection operator $\hat {P}_{m}=|\phi _{m}\rangle \langle \phi _{m}|$,

$$\begin{array}{@{}rcl@{}} {\rho_{S}^{m}} & =&{\Lambda}_{m} (\rho_{S}) \\ & =& \sum\limits_{k}\langle e_{k}| \hat{P}_{m}\hat{U}\rho_{S}\otimes |\tilde{e}_{0}\rangle\langle \tilde{e}_{0}| \hat{U}^{\dag}\hat{P}_{m}^{\dag}|e_{k}\rangle \\ & =& \langle\phi_{m}|\hat{U}|\tilde{e}_{0}\rangle\rho_{S}\langle \tilde{e}_{0}| \hat{U}^{\dag}|\phi_{m}\rangle{\sum}_{k}\langle k|\phi_{m}\rangle\langle\phi_{m}|k\rangle \\ & =& \hat{M}_{m}\rho_{S} \hat{M}_{m}^{\dag} \end{array} $$

(E2)

where $\hat {M}_{m}=\langle \phi _{m}|\hat {U}|\tilde {e}_{0}\rangle$ is the operator defined on $\mathcal {H}_{S}$. The probability of finding outcome m is $p_{m}=Tr({\Lambda }_{m} (\rho _{S}))=Tr(\hat {M}_{m} \hat {M}_{m}^{\dag }\rho _{S})$.

It is evident that if we replace the environment system E with the apparatus system A, the OQS theory gives the same formulations as shown in Section 3. In the case of initial product state, (21) versus (E2) are effectively the same. Let’s consider the case of initial entangled state in the OQS context. Denote the initial system plus environment state as pure bipartite state |Ψ₀〉. After the global unitary operation $\hat {U}$ and subsequent projection $I^{S}\otimes \hat {P}_{m}^{E}=I^{S}\otimes |\phi _{m}\rangle \langle \phi _{m}|$, the composite state becomes $|{\Psi }_{1}\rangle =(I^{S}\otimes \hat {{P_{m}^{E}}})\hat {U}|{\Psi }_{0}\rangle$, take the partial trace over E, we get

$$\begin{array}{@{}rcl@{}} {\rho^{S}_{m}} & =&Tr_{E}(|{\Psi}_{1}\rangle\langle{\Psi}_{1}| \\ & =& \langle\phi_{m}|\hat{U}_{SA}|{\Psi}_{0}\rangle\langle{\Psi}_{0}|\hat{U}_{SA}^{\dag}|\phi_{m}\rangle \\ & =& |\psi_{m}\rangle\langle\psi_{m}|. \end{array} $$

(E3)

This implies the resulting state for S is $|\psi _{m}\rangle =\langle \phi _{m}|\hat {U}_{SA}|{\Psi }_{0}\rangle$, which is equivalent to (28). Taking a similar approach in deriving (29), we can express |ψ_m〉 in the eigenbasis derived from the Schmidt decomposition of |Ψ₀〉. The result is $|\psi _{m}\rangle = 1/\sqrt {p^{\prime }_{m}} {\sum }_{i} \hat {M}_{mi}|\tilde {s}_{i}\rangle$, where the definitions of $\hat {M}_{mi}$ and $p^{\prime }_{m}$ are the same as those in (29) except replacing the apparatus system A with the environment system E. The reduced density operator for S, ${\rho ^{S}_{m}}$, is given by

$$ {\rho_{S}^{m}} = \frac{1}{p^{\prime}_{m}}\sum\limits_{ij}\hat{M}_{mi}|\tilde{s}_{i}\rangle\langle \tilde{s}_{j}|\hat{M}_{mj}^{\dag} $$

(E4)

Equation (E4) can be considered as a generalization of (E2) when the initial state is entangled. There is no simple form of map Λ_m such that ${\rho _{S}^{m}}={\Lambda }_{m}(\rho _{S})$ where ρ_S is the initial density matrix of S. A different representation of ${\rho _{S}^{m}}$ can be derived by rewriting the initial state ρ_SE = ρ_S ⊗ ρ_E + ρ_corr where ρ_corr is a correlation term [25].

Appendix F: Mutual Information

Entanglement between the two systems is measured by the parameter E(ρ) as defined by (13). In Section 3, we also use the mutual information variable to measure the information exchange between the measuring system and the measured system. The mutual information between S and A is defined as I = H(ρ_S) + H(ρ_A) − H(ρ_SA), where H(ρ_SA) is the von Neumann entropy of the composite system S + A. For a composite system S + A that is described by a single relational matrix R, these two variables differ only by a factor of two. However, for a composite system of S + A that is described by an ensemble of relational matrices, the two variables can be very different. This is illustrated by two examples described below.

Case 1

S + A is in an entangled pure state described by $|{\Psi }\rangle _{SA}={\sum }_{i}\lambda _{i}|s_{i}\rangle |a_{i}\rangle$ in Schmidt decomposition, where {λ_i} are the Schmidt coefficients. Subsystem S is in a mixed state. The entanglement measure is $E(\rho _{S})= -{\sum }_{i}{\lambda _{i}^{2}}ln({\lambda _{i}^{2}})$ and the mutual information is $I=-2{\sum }_{i}{\lambda _{i}^{2}}ln({\lambda _{i}^{2}})$.

Case 2

S + A is in a mixed state described by $\rho _{SA}={\sum }_{i}{\lambda _{i}^{2}}|s_{i}\rangle |a_{i}\rangle \langle s_{i}|\langle a_{i}|$. In the case, $H(\rho _{S})=H(\rho _{SA})=H(\rho _{diag})=-{\sum }_{i}{\lambda _{i}^{2}}ln({\lambda _{i}^{2}})$. ρ_SA is a separable bipartite state [20, 21]. There is no entanglement but there is mutual information since $I=-{\sum }_{i}{\lambda _{i}^{2}}ln({\lambda _{i}^{2}})$. Essentially the composite system is a mixed ensemble of product states $\{{\lambda _{i}^{2}}, |s_{i}\rangle |a_{i}\rangle \}$. One can infer that S is in |s_i〉 from knowing A is in |a_i〉, however such mutual information is due to classical correlation. The probability of finding S in an eigenvector |s_i〉 is just the classical probability ${\lambda _{i}^{2}}$.

Although the reduced density operator for S, $\rho _{S} = {\sum }_{i}{\lambda _{i}^{2}}|s_{i}\rangle \langle s_{i}|$, is the same in Case 1 and Case 2, the mutual information is different. More information is encoded in the pure bipartite state in Case 1. When S + A is described by $|{\Psi }\rangle _{SA}={\sum }_{i}\lambda _{i}|s_{i}\rangle |a_{i}\rangle$, besides the inference information between S and A, there is additional indeterminacy due to the superposition at the composite system level. For instance, one cannot determine the composite system S + A is in |s₀〉|a₀〉 or |s₁〉|a₁〉 before measurement. More indeterminacy before measurement means more information can be gained from measurement. This also explains that in the EPR experiment, when Alice measures particle α, she does not only gain information about α, but also gain information about the composite system. Thus, she can predict the state of particle β. On the other hand, such indeterminacy does not exist when S + A is described by a mixture of product state as in Case 2. Since such indeterminacy is for the composite system as a whole, the reduced density operator for a subsystem S, ρ_S, cannot reflect the difference, therefore it appears the same in Case 1 and Case 2.

These two examples show that mutual information variable can substitute the entanglement measurement only when the composite system S + A is described by a single relational matrix R. To quantify change of quantum correlation during a measurement, the entanglement measurement is a more appropriate parameter.